Question

Draining a tank

A cylindrical water tank has a valve at the bottom. When the valve is opened, the water drains out, and rate at which the water flows out is proportional to the fill level because the pressure in the tank is proportional to the mass of water in the tank.

Therefore the decrease of mass in the tank is given by $\frac{dm}{dt} =-\alpha m$ , where m is the mass of water in the tank and α is a constant.

(a) Integrate this differential equation by separation of variables to find the equation that gives the mass of water as a function of time. Assume that the mass of water at t=0 is $m_{0}$.

(b) Assume that the initial mass is 800 kg. You observe that 100 s after opening the vale the remaining mass is 500 kg. How long will you have to wait after opening the valve to that only 50 kg are left?

Answer 1

Dm/dt = -alpha m implies dm/m = -alpha dt implies integral^m_m_0 dm/m = -alpha integral^t_0 dt [therefore where m_0 is initial quantity] implies ln m/m_0 = -alpha t implies m = m_0 e^- alpha t m_0 = 800 kg t = 100 s m = 500 kg implies Ln 500/800 = -alpha times 100 implies alpha = 4.7 times 10^-3 m = 50 kg, m_0 = 800 kg implies t = ln (m/m_0)/- alpha = ln (50/800)/ - 4.7 times 10^-3 sec = 589.8 sec almostequalto 5.90 sec